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16.6 Applying the Derivative

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The derivative can tell us all sorts of things in regards to physical situations. One of these is the instantaneous rate of change. Def: The instantaneous rate of change of y = f (x) per unit change in x at x 0 is: (aka: the derivative evaluated at x 0 ) Ex 1) Physics: Find the velocity and the acceleration of a moving particle with the given position function (meters) at the given time (secs). f (t) = t 3 – 5t 2 + t + 2 at t = 4 v(t) = f (t) = 3t 2 – 10t + 1 f (4) = 3(4) 2 – 10(4) + 1 = 48 – 40 + 1 = 9 m/s a(t) = v (t) = f (t) = 6t – 10 f (4) = 6(4) – 10 = 14 m/s 2

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You don’t always have to find instantaneous rates with respect to time. Ex 2) A spherical balloon is being inflated. Let x represent the radius. Find the instantaneous rate of change of the volume when the radius is 2 ft. (Find deriv. of volume) Volume of sphere: per one foot radius

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We can also use the derivative to find particular maximums or minimums. (Remember the first derivative sign chart helps us determine max’s & min’s) Ex 3) An oil refiner has 100,000 gallons of gasoline that could be sold now with a profit of 25 cents/gal. For each week of delays, an additional 10,000 gal can be produced. However, for each such week, the profit decreases by 1 cent/gal. It is possible to sell all the gasoline that is on hand at any time. When should the gasoline be sold to maximize the profit? *First, we need an equation! Profit function is (gallons sold)(dollars profit per gallon) w = weeks delayed(100,000 + 10,000w)(.25 –.01w) nowaddtl per week nowdecrease per week

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Ex 3) cont… P(w) = (100,000 + 10,000w)(.25 –.01w) = 25,000 – 1000w + 2500w – 100w 2 P(w) = – 100w 2 + 1500w + 25,000 Now, the derivative to find max P (w) = – 200w + 1500 = 0 – 200w = –1500 w = 7.5 P (w) 7.5 –+ rising to falling is a max! Sell right at 7 ½ weeks for max profit

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Homework #1606 Pg 884 #1, 7, 13, 16, 18, 20, 25, 26

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Ch. 5 – Applications of Derivatives 5.6 – Related Rates.

Ch. 5 – Applications of Derivatives 5.6 – Related Rates.

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